Google中國2015校園招聘筆試Round D APAC Test Problem B. GBus count

Problem

There exists a straight line along which cities are built.

Each city is given a number starting from 1. So if there are 10 cities, city 1 has a number 1, city 2 has a number 2,... city 10 has a number 10.

Different buses (named GBus) operate within different cities, covering all the cities along the way. The cities covered by a GBus are represented as 'first_city_number last_city_number' So, if a GBus covers cities 1 to 10 inclusive, the cities covered by it are represented as '1 10'

We are given the cities covered by all the GBuses. We need to find out how many GBuses go through a particular city.

Input

The first line contains the number of test cases (T), after which T cases follow each separated from the next with a blank line.
For each test case, 
The first line contains the number of GBuses.(N) 
Second line contains the cities covered by them in the form 
a₁ b₁ a₂ b₂ a₃ b₃...a_n b_n 
where GBus1 covers cities numbered from a₁ to b₁, GBus2 covers cities numbered from a₂ to b₂, GBus3 covers cities numbered from a₃ to b₃, upto N GBuses. 
Next line contains the number of cities for which GBus count needs to be determined (P).
The below P lines contain different city numbers. 

Output

For each test case, output one line containing "Case #T_i:" followed by P numbers corresponding to the number of cities each of those P GBuses goes through. 

Limits

1 <= T <= 10 
a_i and b_i will always be integers.

Small dataset

1 <= N <= 50 
1 <= a_i <= 500, 1 <= b_i <= 500 
1 <= P <= 50

Large dataset

1 <= N <= 500 
1 <= a_i <= 5000, 1 <= b_i <= 5000 
1 <= P <= 500

Sample

Input
2 4 15 25 30 35 45 50 10 20 2 15 25 10 10 15 5 12 40 55 1 10 25 35 45 50 20 28 27 35 15 40 4 5 3 5 10 27
Output
Case #1: 2 1 Case #2: 3 3 4

Explanation for case 1:
2 GBuses go through city 15 (GBus1 [15 25] and GBus4 [10 20]) 

1 GBus goes through city 25 (GBus1 [15 25]) 

類型：線段樹  難度：2

題意：在一條直線上，有若干城市，城市序號爲順序。給出一些公交線路，以及每條線路覆蓋的起始城市序號和終止城市序號[ai,bi]。再給出一些查詢，每個查詢爲一個城市序號，求這個城市有多少條公交線路經過。

分析：線段樹模板題。由於城市數量爲[1,5000]，所以適合用線段樹解決。需要用到線段樹的成段更新，在給出每條公交線路時，將[ai,bi]區間內+1。查詢時，返回查詢點的值。

代碼如下：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdint>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1

const int N = 5010;
const int maxN = 5010;
int add[N<<2];
int sum[N<<2];

void PushUp(int rt) {
	sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m) {
	if (add[rt]) {
		add[rt<<1] += add[rt];
		add[rt<<1|1] += add[rt];
		sum[rt<<1] += add[rt] * (m - (m >> 1));
		sum[rt<<1|1] += add[rt] * (m >> 1);
		add[rt] = 0;
	}
}

void build(int l,int r,int rt) {
	add[rt] = 0;
	if (l == r) {
		sum[rt] = 0;
		return ;
	}
	int m = (l + r) >> 1;
	build(lson);
	build(rson);
	PushUp(rt);
}

void update(int L,int R,int c,int l,int r,int rt) {		//³É¶ÎÔö¼õ
	if (L <= l && r <= R) {
		add[rt] += c;
		sum[rt] += c * (r - l + 1);
		return ;
	}
	PushDown(rt , r - l + 1);
	int m = (l + r) >> 1;
	if (L <= m) update(L , R , c , lson);
	if (m < R) update(L , R , c , rson);
	PushUp(rt);
}

int srch(int p,int l,int r,int rt) {
	if (l == r) {
		return sum[rt];
	}
	PushDown(rt , r - l + 1);
	int m = (l + r) >> 1;
	if (p <= m) srch(p , lson);
	else srch(p , rson);
}


int main() {
    freopen("B-large.in", "r", stdin);
    freopen("B-large.out", "w", stdout);
    
    int t;
    scanf("%d",&t);

    for(int cnt=1; cnt<=t; ++cnt)
    {
        int n,p;
        build(1,maxN,1);
        scanf("%d",&n);
        for(int i=0; i<n; ++i)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            update(a,b,1,1,maxN,1);
        }
        
        scanf("%d",&p);
        printf("Case #%d:",cnt);
        for(int i=0; i<p; ++i)
        {
            int q;
            scanf("%d",&q);
            printf(" %d",srch(q,1,maxN,1));
        }
        printf("\n");
    }
}